I am trying to solve exercise 1.2 in Aluffi's text "Chapter 0."
Prove that if $\sim$ is a relation on a set $S$, then the corresponding family $\mathscr{P}_{\sim}$ defined in $\S 1.5$ is indeed a partition of $S$: that is, its elements are nonempty, disjoint, and their union is $S$.
Here is my attempt.
By definition, we have $\mathscr{P}_{\sim} = \{[a]_{\sim} \mid a \in S\}$. Each $[a]_{\sim} \in \mathscr{P}_{\sim}$ is nonempty since $a \in [a]_{\sim}$. They are disjoint. Indeed, if $[a]_{\sim} \neq [b]_{\sim}$ are two distinct elements of $\mathscr{P}_{\sim}$, I claim they are disjoint. If $x \in [a]_{\sim} \cap [b]_{\sim}$, then $x \sim a$ and $x \sim b$, so $a \sim x$ by symmetry, and hence $a \sim b$ by transitivity. So if $y \sim a$, $y \sim b$, and if $y \sim b$, then $y \sim a$, so we have $[a]_{\sim} = [b]_{\sim}$. Finally, I claim that $\bigcup\limits_{[a]_{\sim} \in \mathscr{P}_{\sim}} [a]_{\sim} = S$. Clearly, $\bigcup\limits_{[a]_{\sim} \in \mathscr{P}_{\sim}} \subset S$ since $[a]_{\sim} \subset S$ for all $[a]_{\sim} \in \mathscr{P}_{\sim}$. Conversely, if $s \in S$, then $s \in [s]_{\sim} \in \mathscr{P}_{\sim}$, so $s \in \bigcup\limits_{[a]_{\sim} \in \mathscr{P}_{\sim}}$, giving the opposite inclusion. Therefore, the quotient of $S$ by $\sim$ gives a partition of $S$.
How does this look? In particular, have I used the terminology of quotienting by $\sim$ correctly?